第一篇题解的C++实现 | #ranko的手表#

第一篇题解的C++实现

#include <bits/stdc++.h>

#define x first
#define y second

using namespace std;

typedef long long LL;
typedef long double LD;
typedef pair<int, int> PII;

const int N = 1e5 + 10, MOD = 998244353;
const LL INF = 1e18;

void solve() {
    string a, b;
    vector<int> v1, v2;
    cin >> a >> b;

    for (int i = 0; i < 60 * 24; ++i) {
        int h = i / 60;
        int m = i % 60;
        if ((a[0] == '?' || (a[0] - '0') == h / 10) &&
             (a[1] == '?' || (a[1] - '0') == h % 10) &&
             (a[3] == '?' || (a[3] - '0') == m / 10) &&
             (a[4] == '?' || (a[4] - '0') == m % 10))
            v1.push_back(i);

        if ((b[0] == '?' || (b[0] - '0') == h / 10) &&
             (b[1] == '?' || (b[1] - '0') == h % 10) &&
             (b[3] == '?' || (b[3] - '0') == m / 10) &&
             (b[4] == '?' || (b[4] - '0') == m % 10))
            v2.push_back(i);
    }

    int maxd = 0, mind = 60 * 24;
    for (int x : v1) {
        for (int y : v2) {
            if (x < y) {
                maxd = max(maxd, abs(x - y));
                mind = min(mind, abs(x - y));
            }
        }
    }

    cout << mind << ' ' << maxd << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T = 1;
    while (T--) solve();

    return 0;
}